*x* + 5

4*x*2 − 9

 ÷ 

*x*2 + 6*x* + 5

2*x*2 + 3*x* − 9

The given expression is:

\[
\frac{(x + 5)(4x^2 - 9)}{(x^2 + 6x + 5)(2x^2 + 3x - 9)}
\]

### Step 1: Factorizing the components
Let's first factor each part of the expression:

#### Numerator:
1. \(x + 5\) is already factored.
2. \(4x^2 - 9\) is a difference of squares: 
   \[
   4x^2 - 9 = (2x - 3)(2x + 3)
   \]

#### Denominator:
3. \(x^2 + 6x + 5\) factors as:
   \[
   x^2 + 6x + 5 = (x + 5)(x + 1)
   \]
4. \(2x^2 + 3x - 9\) factors as:
   \[
   2x^2 + 3x - 9 = (2x - 3)(x + 3)
   \]

### Step 2: Substitute factored terms into the expression
Now, we rewrite the entire expression with the factored terms:
\[
\frac{(x + 5)(2x - 3)(2x + 3)}{(x + 5)(x + 1)(2x - 3)(x + 3)}
\]

### Step 3: Simplify by canceling common factors
The common factors in the numerator and denominator are \( (x + 5), (2x - 3), \) and \( (x + 3) \). Canceling these gives:
\[
\frac{2x + 3}{x + 1}
\]

### Final Answer:
The simplified expression is:
\[
\frac{2x + 3}{x + 1}
\]

Let me know if you'd like more details on the factorization or the simplification steps.

### Follow-up Questions:
1. How would you factor a trinomial like \(x^2 + 6x + 5\)?
2. Can you explain why \(4x^2 - 9\) is a difference of squares?
3. How do you determine which terms can be canceled when simplifying fractions?
4. What happens to the expression if \(x = -1\)? Why is this important?
5. Can all quadratics be factored this way?

**Tip**: Always check for common factors first when simplifying complex fractions—it often makes the problem easier to solve!

*x* + 5

4*x*2 − 9

÷

*x*2 + 6*x* + 5

2*x*2 + 3*x* − 9

This problem involves simplifying a rational expression by factoring polynomials. Learn how to apply algebraic concepts like the difference of squares and factoring trinomials to simplify the expression (x + 5)(4x^2 - 9) ÷ (x^2 + 6x + 5)(2x^2 + 3x - 9). Suitable for students in Grades 9-12.

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